Given two random variables $X$ and $Y$ , I wish to find $Cov(X + Y, X − Y )$ assuming that

$(a)$ $X$ and $Y$ are independent and

$(b)$ $X$ and $Y$ are dependent & $Var(X) = Var(Y )$

I started with $$Cov(X + Y, X − Y) = Cov(X, X-Y) + Cov(Y, X-Y) = Var(X)- Cov(X,Y) + Cov(Y,X)- Var(Y)$$ such that for $a)$ we have $Var(x) - Var (Y)$ and for $b)$ we have $0$.

Just checking if my method is correct thanks!

  • $\begingroup$ Okay. That's better. $\endgroup$ – Graham Kemp Aug 28 '17 at 1:56
  • $\begingroup$ @xiangqi I believe it is correct. $\endgroup$ – tintinthong Aug 28 '17 at 2:00

Indeed. $~\mathsf {Cov}(X+Y,X-Y)= \mathsf {Var}(X)-\mathsf {Var}(Y)$ whether or not $X,Y$ are independent.  It is a property of the bilinearity of covariance.

Of course, this equals zero when the variances are equal.


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